What is the tangent of f(x)=3+8ln(x),\at x=1 ?

The tangent of f(x)=3+8ln(x),\at x=1 is y=8x-5

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tangent of f(x)=3+8ln(x),\at x=1

tangent of

f (x) = 3 + 8 ln (x), at x = 1

y = 8 x - 5

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = 3 + 8 ln (x) : \frac{df ( x )}{d x} = \frac{8}{x}

Compute the slope of

f (x) = 3 + 8 ln (x)

x = 1 : m = 8

Find the line with slope m=

8

and passing through

(1, 3) : y = 8 x - 5

y = 8 x - 5

(2 x^{3}) (4 x^{2})

\int_{0}^{7} 4 x + 5 - x d x

f (x) = \frac{x ^{2} - 4}{x}

\int 1 + \frac{1}{4} sin^{2} (x) d x

\int e^{x} e^{x} + 1 d x

What is the tangent of f(x)=3+8ln(x),\at x=1 ?
The tangent of f(x)=3+8ln(x),\at x=1 is y=8x-5